Question: Let $x = \sqrt{\frac{\sqrt{53}}{2} + \frac{3}{2}}.$  There exist unique positive integers $a,$ $b,$ $c$ such that
\[x^{100} = 2x^{98} + 14x^{96} + 11x^{94} - x^{50} + ax^{46} + bx^{44} + cx^{40}.\]Find $a + b + c.$
Solution: We have that $x^2 = \frac{\sqrt{53}}{2} + \frac{3}{2}.$  Then $2x^2 = \sqrt{53} + 3,$ so $2x^2 - 3 = \sqrt{53}.$  Squaring both sides, we get
\[4x^4 - 12x^2 + 9 = 53,\]so $4x^4 = 12x^2 + 44.$  Then $x^4 = 3x^2 + 11.$

Since $x \neq 0,$ we can divide both sides of the given equation by $x^{40},$ to get
\[x^{60} = 2x^{58} + 14x^{56} + 11x^{54} - x^{10} + ax^6 + bx^4 + c.\]Now,
\begin{align*}
x^{60} - 2x^{58} - 14x^{56} - 11x^{54} &= x^{54} (x^6 - 2x^4 - 14x^2 - 11) \\
&= x^{54} ((x^2 - 2) x^4 - 14x^2 - 11) \\
&= x^{54} ((x^2 - 2)(3x^2 + 11) - 14x^2 - 11) \\
&= x^{54} (3x^4 - 9x^2 - 33) \\
&= 3x^{54} (x^4 - 3x^2 - 11) \\
&= 0.
\end{align*}So, the equation reduces to
\[x^{10} = ax^6 + bx^4 + c.\]We have that
\begin{align*}
x^6 &= x^2 \cdot x^4 = x^2 (3x^2 + 11) = 3x^4 + 11x^2 = 3(3x^2 + 11) + 11x^2 = 20x^2 + 33, \\
x^8 &= x^2 \cdot x^6 = x^2 (20x^2 + 33) = 20x^4 + 33x^2 = 20(3x^2 + 11) + 33x^2 = 93x^2 + 220, \\
x^{10} &= x^2 \cdot x^8 = x^2 (93x^2 + 220) = 93x^4 + 220x^2 = 93(3x^2 + 11) + 220x^2 = 499x^2 + 1023.
\end{align*}Thus, $x^{10} = ax^6 + bx^4 + c$ becomes
\[499x^2 + 1023 = a(20x^2 + 33) + b(3x^2 + 11) + c.\]Then
\[499x^2 + 1023 = (20a + 3b)x^2 + (33a + 11b + c).\]Since $x^2$ is irrational, we want $a,$ $b,$ and $c$ to satisfy $20a + 3b = 499$ and $33a + 11b + c = 1023.$  Solving for $a$ and $b,$ we find
\[a = \frac{3c + 2420}{121}, \quad b = \frac{3993 - 20c}{121}.\]Hence, $c < \frac{3993}{20},$ which means $c \le 199.$  Also, we want $3c + 2420$ to be divisible by 121  Since 2420 is divisible by 121, $c$ must be divisible by 121.  Therefore, $c = 121,$ which implies $a = 23$ and $b = 13,$ so $a + b + c = \boxed{157}.$